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The temperature of 4.0 moles of a gas occupying $5 \mathrm{dm}^3$ at $3.32 \mathrm{bar}$ is $\left(R=0.083 \mathrm{bar} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
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The correct answer is:
50 K
The amount of the gas, $n=4.0 \mathrm{~mol}$
Volume of gas $Y=5 \mathrm{dm}^3$
Pressure of the gas $p=3.32$ bar
$$
R=0.083 \text { bar dm }^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
$$
From the ideal gas equation, we get
$$
p V=n R T \Rightarrow T=\frac{p V}{n R}=(3 \cdot 32 \times 5) /(4 \times 0 \cdot 083)
$$
Hence, required temperature is $=50 \mathrm{~K}$
Volume of gas $Y=5 \mathrm{dm}^3$
Pressure of the gas $p=3.32$ bar
$$
R=0.083 \text { bar dm }^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
$$
From the ideal gas equation, we get
$$
p V=n R T \Rightarrow T=\frac{p V}{n R}=(3 \cdot 32 \times 5) /(4 \times 0 \cdot 083)
$$
Hence, required temperature is $=50 \mathrm{~K}$
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